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Before introducing a stronger hypothesis, which allows to treat significant cases, we need some notation. We write every element r of R as 1 v 1 v 1 v (1) r = = = = µR(v), 0 π(r) 0 τR(v) 0 In + δR(v) n n n where µR : F AGLn(F), τR : F GLn(F) and δR := τR id : F Matn(F). → → − → The hypothesis we introduce is that δR is linear. First of all, if R is abelian, then δR is linear (see [1]). Moreover, if δR is linear, then R is unipotent by Theorem 2.4(b), but not necessarily abelian. One further motivation for this hypothesis is that δR is linear if and only if = FIn + R is a split local subalgebra of L +1 Matn+1(F). Moreover, two regular subgroups R1 and R2, with δRi linear, are conjugate in AGLn(F) if and only if the corresponding algebras 1 and 2 are isomorphic (see Section 3). In particular, there is a bijection betweL en conjugacyL classes of abelian regular subgroups of AGLn(F) and isomorphism classes of abelian split local algebras of dimension n + 1 over F. This fact was first observed in [1]. It was studied also in connection with other algebraic structures in [3, 4, 5, 6] and in [2], where the classification of nilpotent associative algebras given in [7] is relevant. In Section 7 we classify, up to conjugation, certain types of regular subgroups U of AGLn(F), for n 4, and the corresponding algebras. More precisely, for n = 1 the only regular≤ subgroup is the translation subgroup , which is standard. T For n = 2, 3, we assume that δU is linear. If n = 2 all subgroups U, over any F, are standard (Table 1). For n = 3 the abelian regular subgroups are described in Table 2. The non abelian ones are determined in Lemmas 7.2 and 7.3: there are F conjugacy classes when char F = 2, F + 1 otherwise (see also [7]). If n = 4 |we| assume that U is abelian. The conjugacy| | classes, when F has no quadratic extensions, are shown in Tables 3 and 4. In particular, by the reasons mentioned above, we obtain an independent clas- sification of the split local algebras of dimension n 4 over any field. We obtain also the classification of the commutative split local≤ algebras of dimension 5 over fields with no quadratic extensions. Actually, regular subgroups arise from matrix representations of these algebras. In the abelian case and using the hypothesis that F is algebraically closed, the same classification, for n 6, was obtained by Poonen [12], via commutative algebra. Namely he presents≤ the algebras as quotients of the polynomial ring F[t1,...,tr], r 1. The two approaches are equivalent but, comparing our results with those of≥ Poonen, we detected an inac- curacy1. Namely for n = 5 and char F = 2, the two algebras defined, respectively, by F[x,y,z]/ x2,y2,xz,yz,xy + z2 and F[x,y,z]/ xy,xz,yz,x2 + y2, x2 + z2 are isomorphic (seeh Remark 7.6). i h i Our methods, based on linear algebra, are computer independent.
2. Some basic properties and examples of regular subgroups
Lemma 2.1. A regular submonoid R of AGLn(F) is a subgroup.
n 1 Proof. For any v F , the first row of µR vτR(v)− µR(v) is the same as the ∈ − 1 first row of In+1. From the regularity of R it follows µR vτR(v)− µR(v)= In+1, 1 1 − whence µR(v)− = µR vτR(v)− R. Thus R is a subgroup. − ∈ 1See [13] for a revised version of [12]. MORE ON REGULAR SUBGROUPS OF THE AFFINE GROUP 3
n If δR HomZ (F , Matn(F)), i.e. δR is additive, direct calculation gives that a ∈ regular subset R of AGLn(F) containing In+1 is a submonoid, hence a subgroup, if and only if: n (2) δR(vδR(w)) = δR(v)δR(w), for all v, w F . ∈ Also, given v, w Fn, ∈ (3) µR(v)µR(w)= µR(w)µR(v) if and only if vδR(w)= wδR(v). For the next two theorems we need a basic result, that we recall below for the reader’s convenience. A proof can be found in any text of linear algebra.
Lemma 2.2. Let g GLm(F) have characteristic polynomial χg(t) = f (t)f (t). ∈ 1 2 If (f1(t),f2(t)) = 1, then g is conjugate to h = diag(h1,h2), with χhi (t) = fi(t), i =1, 2. Also, CMatm(F)(h) consists of matrices of the same block-diagonal form.
Lemma 2.3. Let z be an element of a regular subgroup R of AGLn(F). If z is not unipotent then, up to conjugation of R under AGLn(F), we may suppose that
1 w1 0 (4) z = 0 A 0 , 1 0 0 A2 where A1 is unipotent and A2 does not have the eigenvalue 1. 1 w Proof. Let z = , where z = I + δ (w). Since z has the eigenvalue 1 [14, 0 z 0 n R 0 0 Lemma 2.2], the characteristic polynomial χz0 (t) of z0 factorizes in F[t] as
m1 χz (t) = (t 1) g(t), g(1) =0, m 1, deg g(t)= m 1. 0 − 6 1 ≥ 2 ≥ By Lemma 2.2, up to conjugation in GLn(F), we may set z0 = diag(A1, A2) with:
m1 χA (t) = (t 1) , χA (t)= g(t). 1 − c 2 Write w = (w , w ) with w Fm1 and w Fm2 . Since A does not have the 1 2 1 ∈ 2 ∈ 2 eigenvalue 1, the matrix A2 Im2 is invertible. Conjugating by the translation 1 − 1 0 w (A Im )− 1 w 0 2 2 − 2 1 0 Im 0 we may assume w = 0, i.e. z = 0 A 0 . 1 2 1 0 0 Im2 0 0 A2 n We observe that if δR HomF (F , Matn(F)), i.e. δR is linear, then there is a ∈ natural embedding of R into a regular subgroup R of AGLn(F) for any extension F of F. Namely: b b 1v ˆ n b R = :v ˆ F F F AGLn(F), 0 In + δ b(ˆv) ∈ ⊗ ≤ R b b b where δ b λvˆ i = λˆiδR(vi), λˆi F. Clearly δ b is linear. R i i ∈ R We obtain the following consequences. P P b Theorem 2.4. Let R be regular subgroup of AGLn(F). Then the following holds: (a) the center Z(R) of R is unipotent; (b) if δR is linear, then R is unipotent. 4 M.A.PELLEGRINIANDM.C.TAMBURINIBELLANI
Proof. (a) Our claim is clear if Z(R) = 1 . So let 1 = z Z(R) and assume, by contradiction, that z is not unipotent.{ By} Lemma 2.36 up∈ to conjugation z has shape (4). Lemma 2.2 gives the contradiction that its centralizer is not transitive on the affine vectors. (b) Let F be the algebraic closure of F. By what observed above, substituting R with R, if necessary, we may assume F algebraically closed. By contradic- tion, supposeb that z R is not unipotent. Up to conjugation we may sup- ∈ pose z asb in the statement of Lemma 2.3. In the same notation, δR(w1, 0) = diag(A Im , A Im ). Let ξ F be an eigenvalue of A Im and 0 = 1 − 1 2 − 2 ∈ 2 − 2 6 v Fm2 be a corresponding eigenvector. We have ξ = 0 and, by linearity, ∈ 1 1 1 6 δR( ξ− w1, 0) = diag ξ− (A1 Im1 ) , ξ− (A2 Im2 ) . It follows that the − − − 1− − first row of the product µR(0, v)µR( ξ− w1, 0) is equal to the first row of the 1 − second factor µR( ξ− w1, 0). From the regularity of R we get µR(0, v)= In+1. In particular v = 0, a− contradiction. We conclude that R is unipotent.
The following examples show how the linearity of δR seems to be necessary to manage a classification of the regular subgroups of AGLn(F), even when unipotent. Example 2.5. Let be a basis of R over Q. For every subset S of denote by B B fS : R the function such that fS(v) = v if v S, fS(v) = 0 otherwise. B → ∈ Let fS be its extension by linearity to R. In particular fS HomZ(R, R) and ∈ Im (fS) = SpanR(S), the subspace generated by S. Considerb the regular subgroup RS of the affine group AGLb 2(R) defined by: b 1 x y b f (y) RS = 0 e S 0 : x, y R . ∈ 0 0 1 r The set of eigenvalues of the matrices in RS is e : r SpanR(S) . It follows that { ∈ } S = S gives RS not conjugate to RS under GL (R). A fortiori RS and RS are 1 6 2 1 2 3 1 2 not conjugate under AGL2(R). Thus in AGL2(R) there are as many non conjugate R regular subgroups as possible, namely 2| |.
Example 2.6. Let F = Fp be a field of characteristic p> 0. Then the set 6 1 x1 x2 R = 0 1 xp : x , x F 1 1 2 ∈ 0 0 1 F is a non abelian unipotent regular subgroup of AGL2( ) such that δR is not linear.
Example 2.7 (Heged˝us, [8]). Let n 4 and F = Fp (p odd). Take the matrices ≥ 1 2 2 − A = diag(A3, In 4) and J = diag(J3, In 4) of GLn 1(Fp), where A3 = 0 1 2 − − − 0 0− 1 0 0 1 and J3 = 0 1 0 . Then A has order p, its minimum polynomial has degree 3 and 1 0 0 AJAT = J. The subset R of order p4, defined by vJvT 1 v h + 2 h h T n 1 R = 0 A A Jv : v Fp − ,h Fp , ∈ ∈ 00 1 is a regular submonoid of AGLn(Fp), hence a regular subgroup. Moreover R = ∩ T 1 . Note that δR is not linear: suppose the contrary and observe that the elements { } MORE ON REGULAR SUBGROUPS OF THE AFFINE GROUP 5
µR(vn) and µR(2vn) correspond to h = 1 and h = 2 respectively (vn = (0,..., 0, 1)). 2 Thus, in order that δR(2vn)=2δR(vn) we should have A In 1 = 2A 2In 1, 2 − − − − i.e. (A In 1) = In 1, against our choice of A. − − − 3. Algebras and regular subgroups In this section we highlight the connections between regular subgroups and finite dimensional split local algebras over a field F. To this purpose we recall that an F-algebra with 1 is called split local if /J( ) is isomorphic to F, where J( ) denotes theL Jacobson radical of . In particularL L = F 1+ J( ), where the setL F 1 = α1 : α F is a subringL of containedL in its centreLZ( ). Note that { L ∈ } L L J( ) is the set ∗ of the invertible elements of . If J( ), viewed as an F- module,L\ L has finite dimension,L we say that is finite dimensionalL L . If ψ is an isomorphism between two localL split F-algebras , , then ψ(J( )) = L1 L2 L1 J( 2) and ψ(α 1 )= α1 for all α F. L L1 L2 ∈ Theorem 3.1. Let be a finite dimensional split local F-algebra. In the above L notation, set n = dimF(J( )). Then, with respect to the product in , the subset L L R =1+ J( )= 1+ v : v J( ) L { ∈ L } is a group, isomorphic to a regular subgroup of AGLn(F) for which δR is linear. Proof. Clearly R is closed under multiplication. Moreover any r R has an inverse 1 1 ∈ in . From r 1 J( ) we get r− 1 J( ), whence r− R. We conclude L − ∈ L − ∈ L ∈ that R is a group. Consider the map R AGLn(F) such that, for all v J( ): → ∈ L 1 v (5) 1+ v B , 7→ 0 In + δR(v) where v and δR(v) are, respectively, the coordinate vector of v and the matrix of the rightB multiplication by v with respect to a fixed basis of J( ), viewed as F-module. In particular, considering the right multiplication byB w: L
(6) (vw) = v δR(w), for all v, w J( ). B B ∈ L The map (5) is injective and we claim that it is a group monomorphism. Set δ = δR for simplicity. For all v, w J( ) we have δ(vw) = δ(v)δ(w) by the associativity law and δ(v + w)= δ(v)+ δ∈(w)L by the distributive laws. Now: 1 v + w + (vw) (1 + v)(1 + w)=1+ v + w + vw B B B . 7→ 0 In + δ(v)+ δ(w)+ δ(vw) On the other hand, considering the images of 1 + v and 1+ w: 1 v 1 w 1 w + v + v δ(w) B B = B B B . 0 In + δ(v) 0 In + δ(w) 0 In + δ(v)+ δ(w)+ δ(v)δ(w) Thus (5) is a homomorphism if and only if (6) holds. This proves our claim. Notice that Theorem 3.1 shows how to construct a regular subgroup starting from the presentation of a split local algebra. Example 3.2. As in [12], consider the split local algebra F[t ,t ,t ] = 1 2 3 . L t 2 + t 2,t 2 + t 2,t t ,t t ,t t h 1 2 1 3 1 2 1 3 2 3i 6 M.A.PELLEGRINIANDM.C.TAMBURINIBELLANI
4 2 In this case J( ) = F has a basis given by t1,t1 ,t2,t3 and considering the L ∼ { } 2 right multiplication by an element of this basis we get δ(t1) = E1,2, δ(t1 ) = 0, δ(t )= E , , δ(t )= E , , and so the associated regular subgroup is 2 − 3 2 3 − 4 2 1 x1 x2 x3 x4 0 1 x 0 0 1 R = 00 1 00 : x , x , x , x F . 1 2 3 4 ∈ 0 0 x3 1 0 − 0 0 x4 0 1 − Conversely we have the following result. n Theorem 3.3. Let R = µR(F ) be a regular subgroup of AGLn(F). Set
0 v n V = R In+1 = v F and R = FIn+1 + V. − 0 δR(v) | ∈ L (a) If δR is additive, then V is a subring (without identity) of Matn+1(F); (b) the function δR is linear if and only if R is a split local subalgebra of L Matn (F) with J( R)= V . +1 L Proof. Set µ = µR, δ = δR, I = In+1 and = R. (a) Since δ is additive, V is an additive subgroup.L L From (µ(v) I)(µ(w) I) = (µ(v)µ(w) I) (µ(v) I) (µ(w) I) it follows that V is a subring.− − − − − − − (b) Suppose first that δ is linear. Using (a) we have that is an additive subgroup. By the linearity, (FI)V = V (FI) = V . It follows thatL = , hence is a LL L L subalgebra of Matn+1(F). Again linearity gives αI + (µ(αv) I) = αµ(v) for all n − α F, v F . Thus V = F∗R consists of elements with inverse in . We conclude∈ that∈ V = J( )L\ and is a split local F-algebra. L Vice-versa, let beL a splitL local subalgebra with J( ) = V . In particular V is an additive subgroup,L whence δ(v + w)= δ(v)+ δ(w) forL all v, w Fn. Since V is an ideal we get δ(αv)= αδ(v) for all α F, v Fn. ∈ ∈ ∈
Our classification of the regular subgroups of AGLn(F) is based on the following proposition (see [1, Theorem 1]), where GLn(F) is defined as in the Introduction. F Proposition 3.4. Assume that R1, R2 arec regular subgroups of AGLn( ) such that δR1 and δR2 are linear maps. Then the following conditions are equivalent:
(a) R1 and R2 are conjugate in GLn(F); (b) R1 and R2 are conjugate in AGLn(F); (c) the algebras R and R arec isomorphic. L 1 L 2 Proof. Set δ = δR , = R , δ = δR , = R , µ = µR and I = In . 1 1 L1 L 1 2 2 L2 L 2 1 1 +1 (a)= (b)= (c) is clear. Let us prove (c)= (a). By Theorem 3.3, 1 and 2 are split local⇒ algebras⇒ with Jacobson radicals R ⇒ I and R I, respectively.L SupposeL 1 − 2 − that ψ : 1 2 is an algebra isomorphism. In particular ψ(R1 I) = R2 I L → L n − − and ψ induces an F-automorphism of F . Let P GLn(F) be the matrix of this automorphism with respect to the canonical basis∈ of Fn. Then 0 v 0 vP ψ = . 0 δ (v) 0 δ (vP ) 1 2 MORE ON REGULAR SUBGROUPS OF THE AFFINE GROUP 7
n For all v, w F we have ψ(µ1(w)µ1(v)) = ψ(µ1(w))ψ(µ1(v)). This implies ∈ 1 n δ (v)P = Pδ (vP ), whence δ (vP )= P − δ (v)P for all v F . We conclude: 1 2 2 1 ∈ 1 0 v 0 vP 1 0 − 0 v 1 0 ψ = = . 0 δ (v) 0 δ (vP ) 0 P 0 δ (v) 0 P 1 2 1
4. Centralizers of unipotent elements
Our classification of unipotent regular subgroups of AGLn(F) is connected to the classical theory of canonical forms of matrices. For the reader’s convenience we recall the relevant facts. For all m 2, the conjugacy classes of unipotent elements ≥ in GLm(F) are parametrized by the Jordan canonical forms
(7) J = diag(Jm1 ,...,Jmk ), m = mi, where each Jmi is a Jordan block of size mi, namely aX matrix having all 1’s on the main diagonal and the diagonal above it, and 0’s elsewhere. A Jordan block Jm has minimal polynomial (t 1)m and its centralizer is the m-dimensional algebra − x0 x1 ... xm 2 xm 1 − − 0 x0 ... xm 3 xm 2 . − .− (8) Tm,m(F) := .. . : xi F ∈ 0 0 ... x x 0 1 0 0 ... 0 x 0 generated by Jm. To study the centralizer of J in (7) we write c Mat m(F) as: ∈ C1,1 ... C1,k . . . (9) c = . . . , Ci,j Matm ,m (F). . . . ∈ i j C ... C k,1 k,k Clearly c centralizesJ if and only if
(10) Jmi Ci,j = Ci,j Jmj , for all i, j.
Lemma 4.1. Take J as in (7) and assume further that m1 mk. Then the element c above centralizes J if and only if ≥···≥
Ci,j Tm ,m (F), for all i, j, ∈ i j where each Tmi,mi (F) is defined as in (8) with m = mi and, for mi >mj:
Tmj ,mj (F) 0 Tm ,m (F) := ,Tm ,m (F) := 0 ... 0 Tm ,m (F) . i j . j i j j . 0 Lemma 4.2. Take J as in (7). If m < mi for some i 2, then the group 1 ≥ CMat (F)(J) AGLm 1(F) is not transitive on affine row vectors (1, x1,...,xm 1). m ∩ − − Proof. Consider c C F (J) and decompose it as in (9). Application of (10) ∈ Matm( ) and elementary matrix calculation give that, whenever m1 < mi, the first row of the matrix C1,i must be the zero vector. Our claim follows immediately. 8 M.A.PELLEGRINIANDM.C.TAMBURINIBELLANI
Lemma 4.3. Let J be as in (7). If k > 1, assume further that m1 mi for all m ≥ i 2. Then, for every v F , there exists c CMatm(F)(J) having v as first row. If≥ the first coordinate of v∈is non zero, such c∈can be chosen nonsingular.
Proof. Both claims are direct consequences of the description of Tm,m(F) preceding (8) and Lemma 4.1.
Theorem 4.4. Let R be a regular subgroup of AGLn(F) and 1 = z be an element 6 of the center Z(R) of R. Then, up to conjugation of R under AGLn(F), we may suppose that z = Jz, where Jz = diag (Jn1 ,...,Jnk ) is the Jordan form of z having Jordan blocks of respective sizes ni ni for all i 1. ≥ +1 ≥ Proof. Let t be the number of non-trivial invariant factors of z. 1 Case t = 1, i.e. Jz = Jn+1. Let g GLn+1(F) be such that g− zg = Jz. Since T ∈ T e0 is the eigenspace of Jz (acting on the left) we have that ge0 must be the T T eigenspace of z (acting on the left). From z AGLn(F) it follows ze0 = e0 , hence T T 1 ∈ ge = λe . We conclude that λ− g AGLn(F) conjugates z to Jz. 0 0 ∈ Case t> 1. By the unipotency of z, there exists g GLn(F) that conjugates z to ∈ 1 w1 ... wh c 0 Jm1 z′ = . . . . .. 0 0 ... J mh ni We claim that the first coordinate of wi F cannot be 0 for all i 1. Indeed, in ni ∈ ≥ this case, there exists ui F such that wi = ui (In Jn ) for all i 1. Setting ∈ i − i ≥ u = (u1,...,uh) we have 1 1 u 1 u − 1 0 z := z = , J = diag (J ,...,J ) . ′′ 0 I ′ 0 I 0 J m1 mh
It follows that CAGLn(F)(z′′) is not transitive on the affine vectors by Lemma 4.2, noting that J = In: a contradiction. So there exists some t, 1 t h, such that 6 ≤ ≤ the first coordinate of wt is non zero. Up to conjugation by an obvious permutation matrix in GLn(F) we may assume t = 1, i.e., wt = w1.
By Lemma 4.3 there exists p GLn(F) which centralizes diag (Jm1 ,...,Jmh ) ∈ 1 and has v =c (w1,...,wh) as first row. It follows that vp− = (1, 0,..., 0). Thus 1 0 1 0 z′′′ = z′ 1 = diag(J1+m1 ,Jm2 ,...,Jmh ). 0 p 0 p− Again by Lemma 4.2 we must have m +1 mi for all i 2. A final conjugation, 1 ≥ ≥ if necessary, by a permutation matrix in GLn(F) allows to arrange the blocks of z′′′ in non-increasing sizes, i.e. allows to conjugate z′′′ to Jz as in the statement. c 5. Some useful parameters We introduce some parameters that will be used mainly to exclude conjugacy among regular subgroups. Let H be any unipotent subgroup of Gn := AGLn(F). For each h In H In we may consider the degree of its minimal polynomial − +1 ∈ − +1 over F, denoted by deg minF(h In ), and its rank, denoted by rk(h In ). E.g., − +1 − +1 deg minF(h In )= n+1 if and only if h is conjugate to a unipotent Jordan block − +1 Jn of size n + 1. Note that rk(Jn In )= n. +1 +1 − +1 MORE ON REGULAR SUBGROUPS OF THE AFFINE GROUP 9
Hence, we may set:
d(H) = max deg minF(h In ) h H ; { − +1 | ∈ } r(H) = max rk(h In+1) h H ; { −n | ∈ } k(H) = dimF w F : wπ(h)= w . { ∈ } If H is a subgroup of a regular subgroup U such that δU is linear, then k(H) = dimF Ker(δU H ). | Clearly if two unipotent subgroups H1, H2 are conjugate, then:
d(H1)= d(H2), r(H1)= r(H2), k(H1)= k(H2),
d(Z(H1)) = d(Z(H2)), r(Z(H1)) = r(Z(H2)), k(Z(H1)) = k(Z(H2)). n Lemma 5.1. Let H be a unipotent subgroup of AGLn(F) and assume that F is the direct sum of non-trivial π(H)-invariant subspaces V1,...,Vs. Then k(H) s. In particular, if k(H)=1, then H is indecomposable. ≥
Proof. H induces on each Vi a unipotent group Hi. So, in each Vi, there exists a non-zero vector wi fixed by all elements of Hi (see [10, Theorem 17.5 page 112]). It n follows that w1,...,ws are s linearly independent vectors of F fixed by π(H). Notation. For sake of brevity, we write 1 x x ... x (11) U = 1 2 n 0 τU (x , x ,...,xn) 1 2 to indicate the regular subgroup
1 x1 x2 ... xn U = : x1,...,xn F . 0 τU (x1, x2,...,xn) ∈ For all i n, we denote by Xi the matrix of U In obtained taking xi = 1 and ≤ − +1 xj = 0 for all j = i. 6 n The set v ,...,vn is the canonical basis of F . { 1 } We recall that the center of a unipotent group is non-trivial.
Lemma 5.2. Let U be a unipotent regular subgroup of Gn. If d(Z(U)) = n +1 then, up to conjugation, U = CGn (Jn+1). Moreover U is abelian.
Proof. Up to conjugation Jn+1 Z(U), whence U CGn (Jn+1). Since this group ∈ ≤ is regular, we have U = CGn (Jn+1). It follows that U is abelian.
Lemma 5.3. Let U be a regular subgroup of Gn such that δU is linear. If r(U)=1, then U is the translation subgroup , which is an abelian normal subgroup of Gn. Furthermore, d( )=2. T T Proof. By the unipotency, we may always assume that U is upper unitriangular. Then, the rank condition gives δ(vi)=0, for 1 i n, whence our claim. ≤ ≤ Lemma 5.4. Let U be a regular subgroup of Gn, n 2, such that δU is linear. If d(Z(U)) = n then, up to conjugation, for some fixed≥α F: ∈ 1 x1 x2 ... xn−2 xn−1 xn 0 1 x1 ... xn−3 0 xn−2 0 0 1 ... xn−4 0 xn−3 . . . . . (12) U = Rα = ...... 0 0 0 ... 1 0 x1 0 0 0 ... 0 1 αxn−1 0 0 0 ... 0 0 1 10 M.A.PELLEGRINIANDM.C.TAMBURINIBELLANI
In particular U is abelian and r(U) = n 1. Furthermore, R0 and Rα are not conjugate for any α =0, − 6 If n 3, an epimorphism Ψ: F[t ,t ] FIn + R is obtained setting ≥ 1 2 → +1 0 (13) Ψ(t1)= X1, Ψ(t2)= Xn 1. − In this case we have Ker(Ψ) = t n,t 2,t t . h 1 1 1 2i If n 4 is even, then Rα is conjugate to R for any α =0 and an epimorphism ≥ 1 6 Ψ: F[t ,t ] FIn + Rα is obtained setting 1 2 → +1 n−2 (14) Ψ(t1)= αX1, Ψ(t2)= α 2 Xn 1. − n 1 2 In this case Ker(Ψ) = t1 − t2 ,t1t2 . h − 2 i If n is odd, write α =0 as α = λε , λ, ε F∗. Then Rα is conjugate to Rλ and 6 ∈ an epimorphism Ψ: F[t1,t2] FIn+1 + Rα is obtained setting → − n 3 n 2 (15) Ψ(t1)= αX1, Ψ(t2)= λ 2 ε − Xn 1. − n 1 2 In this case Ker(Ψ) = t − λt ,t t . In particular Rα and Rβ (α, β F∗) h 1 − 2 1 2i ∈ are conjugate if and only if β/α is a square in F∗.
Proof. Up to conjugation we may suppose that z = diag(Jn,J1) Z(U). The X 0 ∈ subalgebra generated by z coincides with the set : X CG − (Jn) . 0 1 ∈ n 1 This information gives the values of δ(vi) for 1 i n 1. From Lemma 4.1, we ≤ ≤ − get δ(vn) = αEn,n 1. Applying (3), it follows that U is abelian. Conjugating by the permutation matrix− associated to the transposition (n,n + 1) we obtain that U is conjugate to Rα. The subgroups R and Rα are not conjugate when α = 0, since k(R ) = 2 and 0 6 0 k(Rα) = 1. The presentations of the corresponding algebras can be verified by matrix calculation. 2 Assume now that n is odd. If α = βε = 0, then the subgroups Rα and Rβ are 6 conjugate in virtue of (15) and Proposition 3.4. Conversely, suppose that α, β F∗ 1 ∈ and that Q− RαQ = Rβ for some Q GLn(F) (see Proposition 3.4). According to (12), write ∈ c 1 X xn 1 Y yn T T Rα = 0 In 1 + DX AX , Rβ = 0 In 1 + DY BY , 0− 0 1 0− 0 1 where X = (x1,...,xn 1), Y = (y1,...,yn 1), A = diag(antidiag(1,..., 1), α) and B = diag(antidiag(1−,..., 1),β). Hence det(− A) = ζα and det(B) = ζβ, where n+1 ζ = ( 1) 2 . We may assume that vnQ = λvn since vn is the subspace fixed − n 1h i pointwise by both subgroups. Thus, for some N F − and λ F∗, Q = Q1Q2, where ∈ ∈ 10 0 1 0 0 1 Q1 = 0 In 1 λ− N ,Q2 = 0 M 0 , det(M) =0, λ =0. 00− 1 0 0 λ 6 6
1 Now, Q1− RαQ1 = Rα, where
1 X x˜n 1 X x˜n e 1 T T Rα = 0 In 1 + DX λ− DX N + AX = 0 In 1 + DX AX , − − 00 1 0 0 1 e e MORE ON REGULAR SUBGROUPS OF THE AFFINE GROUP 11 with det(A) = ζα. From RαQ2 Q2Rβ we get, in particular, Y = XM and T T T− T T n 1 MBY = λAX , whence MBM X = λAX , for all X F − . It follows that ∈ MBM T =eλA and taking thee determinant of both sides, we obtain ζβ(det(M))2 = n 1 ζλ − α. Wee conclude that β/α is a square.e e 6. Standard regular subgroups The aim of this section is to define, for every partition λ of n + 1 different from n+1 ♯ (1 ), one or two abelian regular subgroups Sλ,Sλ of AGLn(F) so that different partitions define non-conjugate subgroups.
To this purpose we start by identifying the direct product AGLm1 (F) AGLm2 (F) m1 m2 × with the stabilizer of F and F in AGLm1+m2 (F), namely with the subgroup: 1 v w m1 m2 0 A 0 : v F , w F , A GLm (F), B GLm (F) . ∈ ∈ ∈ 1 ∈ 2 0 0 B Clearly, in this identification, if U are respective regular subgroups of AGL (F) i mi for i =1, 2 then U U is a regular subgroup of AGLm m (F). 1 × 2 1+ 2
Here, and in the rest of the paper, we denote by S(1+n) the centralizer in AGLn(F) of a unipotent Jordan block of size n + 1, namely:
S(1+n) = CAGLn(F)(J1+n).
Moreover we write τ1+n for τS(1+n) and δ1+n for δS(1+n) . By Lemma 5.2, a regular subgroup of AGLn(F ) is conjugate to S(1+n) if and only if d(Z(S(1+n))) = n + 1. Observe that S(1+n) is abelian and indecomposable by Lemma 5.1. More generally, for a partition λ of n + 1 such that:
(16) λ = (1+ n ,n ,...,ns), s 1, ni n i 1, 1 i s 1, 1 2 ≥ ≥ 1+ ≥ ≤ ≤ − we define an abelian regular subgroup Sλ of AGLn(F). Namely we set: s
Sλ = S(1+n1,...,ns) = S(1+nj ). j=1 Y n 1 Notice that Sλ is indecomposable only for s = 1. In particular, if λ = (2, 1 − ), then Sλ = . T Lemma 6.1. Given a partition (1 + n ,n ) of n +1, with 1+ n n > 1, set: 1 2 1 ≥ 2 1 u v ♯ T n1 n2 S = 0 τ1+n (u) w Du : u F , v F , (1+n1,n2) 1 ⊗ ∈ ∈ 0 0 τn2 (v) n ♯ where w = (0,..., 0, 1) F 2 , D = antidiag(1,..., 1) GLn (F). Then S 1 (1+n1,n2) ∈ ∈ ♯ is an indecomposable regular subgroup of AGLn(F) with d(S )= n +2. (1+n1,n2) 1 Proof. Routine calculation with matrices shows that S♯ is closed under mul- (1+n1,n2) tiplication, hence a subgroup. Moreover it is indecomposable by Lemma 5.1. Again by matrix calculation one can see that d(S♯ ) = n + 2. To check that (1+n1,n2) 1 S♯ is a subgroup it is useful to note that: (1+n1,n2) (a) all components of u(w DuT), except possibly the last one, are zero; ⊗ 12 M.A.PELLEGRINIANDM.C.TAMBURINIBELLANI
(b) δn2 (v) does not depend on the last component of v; T (c) (w Du )δn2 (v) = 0; ⊗ T T n1 (d) w D(u1δ1+n1 (u2)) = δ1+n1 (u1)(w Du2 ) for all u1,u2 F . ⊗ ⊗ ∈
Next, consider a partition µ of n + 1 such that:
(17) µ = (1+n ,n ,...,ns), s 2, 1+n n > 1, ni n i 1, 2 i s 1. 1 2 ≥ 1 ≥ 2 ≥ 1+ ≥ ≤ ≤ − ♯ We define the abelian regular subgroup Sµ in the following way: s ♯ ♯ ♯ Sµ = S = S S(1+nj ). (1+n1,n2,n3...,ns) (1+n1,n2) × j=3 Y ♯ The regular subgroups Sλ,Sµ associated to partitions as above will be called ♯ standard regular subgroups. As already mentioned Sλ and Sµ are always abelian. n Remark 6.2. Let λ = (1+ n1,...,ns) be a partition as in (16). Then F is a direct sum of s indecomposable modules of respective dimensions ni for which d = ni + 1. Furthermore,
F[t1,t2,...,ts] (18) λ = FIn+1 + Sλ = . ∼ ni+1 L t :1 i s; titj :1 i < j s h i ≤ ≤ ≤ ≤ i An epimorphism Ψ : F[t ,t ,...,ts] λ is obtained by setting 1 2 → L (19) Ψ(ti)= Xi. n Let now µ = (1+ n1,...,ns) be a partition as in (17). Then F is a direct sum of s 2 indecomposable modules of respective dimension ni, i 3, for which − ≥ d = ni + 1, and a single indecomposable module of dimension n1 + n2 for which d = n1 + 2. Furthermore,
♯ ♯ F[t1,t2,...,ts] (20) = FIn+1 + S = . µ µ ∼ n1+1 n2 ni+1 L t t ; t :3 i s; titj :1 i < j s h 1 − 2 i ≤ ≤ ≤ ≤ i ♯ An epimorphism Ψ : F[t ,t ,...,ts] is obtained by setting 1 2 → Lµ (21) Ψ(ti)= Xi.
Theorem 6.3. Let λ1 = (1+n1,...,ns), λ2 = (1+m1,...,mt) be two partitions of n +1 as in (16) and let µ1 = (1+ a1,...,ah),µ2 = (1+ b2,...,bk) be two partitions of n +1 as in (17). Then ♯ (a) Sλ1 is not conjugate to Sµ1 ;
(b) Sλ1 is conjugate in AGLn+1(F) to Sλ2 if and only if λ1 = λ2; ♯ F ♯ (c) Sµ1 is conjugate in AGLn+1( ) to Sµ2 if and only if µ1 = µ2. ♯ Proof. (a) There exists an isomorphism ϕ : λ . Setting vu := vϕ(u) L 1 → Lµ1 J ♯ J ♯ for all u λ1 , v ( µ1 ) we may consider ( µ1 ) as an λ1 -module. Clearly ∈ L ♯ ∈ nL L Ln J( λ ) = J( ) = F as λ -modules. By construction, F is a direct sum of L 1 ∼ Lµ1 ∼ L 1 indecomposable λ1 -modules. By Remark 6.2, for J( λ1 ) each direct summand L ♯ L has dimension ni and d = ni + 1 and for J( ) there is a direct summand of Lµ1 dimension a1 + a2 and d = a1 +2 (b) Arguing as before, we may consider J( λ2 ) as an λ1 -module. In this case, n L L by construction, F is a direct sum of indecomposable λ -modules of dimension L 1 MORE ON REGULAR SUBGROUPS OF THE AFFINE GROUP 13 ni such that d = ni + 1 (see Remark 6.2(1)). By Krull-Schmidt Theorem they are conjugate if and only if s = t and ni = mi. (c) Again, we may consider J( ♯ ) as an ♯ -module. In this case, by construc- Lµ2 Lµ1 tion, Fn is a direct sum of indecomposable ♯ -modules, one of dimension a + a Lµ1 1 2 such that d = a1 +2 Remark 6.4. Note that for n 3, the regular subgroups R0 and R1 of Lemma 5.4 ≥ ♯ coincide, respectively, with S(n,1) and S(n 1,2). − 7. Regular subgroups with linear δ By [14, Lemma 5.1], the only regular subgroup of AGL1(F) is the translation x0 x1 subgroup = S(2) = . However, already for n = 2, a description T 0 x0 R becomes much more complicated. As seen in Example 2.5 there are 2| | conjugacy classes of regular subgroups of AGL2(R) with trivial center. Also, restricting to the unipotent case, one may only say that every unipotent regular subgroup is conjugate to: 1 x1 x2 0 1 σ(x1) 00 1 where σ HomZ(F, F) ([14, Lemma 5.1]). See also Examples 2.6 and 2.7. ∈ Thus, from now on, we restrict our attention to regular subgroups U of AGLn(F) such that the map δU defined in (1) is linear. U is unipotent by Theorem 2.4 and so there exists 1 = z Z(U). By Theorem 4.4, up to conjugation, we may assume that z is a Jordan6 form.∈ For the notation we refer to Section 5. Our classification is obtained working on the parameters d, r and k, considered in this order. In the tables of the next subsections the indecomposability of the regular sub- 4 groups follows from Lemma 5.1, since k(U) = 1, except for the subgroup U1 of Table 3, for which we refer to Lemma 7.4. Also, we describe the kernel of the epimorphism Ψ: F[t ,...,ts] FIn + U. 1 → +1 7.1. Case n = 2. If d(Z(U)) = 3, then U is abelian and conjugate to S(3) by Lemma 5.2. If d(Z(U)) = 2, then U is abelian and r(U) = 1 by Lemma 5.4. Thus U is conjugate S(2,1) by Lemma 5.3. Table 1 summarizes these results. U FIn+1 + U Ψ Ker(Ψ) x0 x1 x2 3 S 0 x0 x1 (19) t indec. (3) h 1 i 0 0 x0 x0 x1 x2 S 0 x 0 (19) t ,t 2 (2,1) 0 h 1 2i 0 0 x0 Table 1. Representatives for the conjugacy classes of regular sub- groups U of AGL2(F) with linear δ, for any field F. 14 M.A.PELLEGRINIANDM.C.TAMBURINIBELLANI 7.2. Case n =3. To obtain the full classification, we need some preliminary results. Lemma 7.1. If d(Z(U)) = r(Z(U)) =2, then U is abelian, conjugate to 1 x1 x2 x3 3 0 1 0 x2 U1 = 0 0 1 x1 00 0 1 and char F =2. An epimorphism Ψ: F[t ,t ] FI + U is obtained by setting 1 2 → 4 (22) Ψ(t1)= X1, Ψ(t2)= X2. In this case, we have Ker(Ψ) = t 2,t 2 . h 1 2 i Proof. We may assume z = diag(J ,J ) Z(U), whence δ(v ) = E , . From 2 2 ∈ 1 2 3 Lemma 4.1 and the unipotency of U we obtain δ(v2) = E1,3 + αE2,1 + βE2,3. It follows v1δ(v2)= v3. Now, we apply (2) to v1, v2, which gives δ(v3)= δ(v1)δ(v2)= 0. Direct calculation shows that U is abelian. Hence d(Z(U)) = d(U) = 2. In 2 2 particular, (µ(v2) I4) = 0 gives α = β = 0. Finally (µ(v1 + v2) I4) = 0 gives char F = 2. − − It is convenient to denote by V (α2, α3,β2,β3) the regular subgroup defined by: 1 x1 x2 x3 0 1 00 (23) V (α2, α3,β2,β3)= . 0 α2x2 + α3x3 1 0 0 β2x2 + β3x3 0 1 Notice that V (α2, α3,β2,β3) is abelian if and only if α3 = β2. T We need the cosquare A− A of a nonsingular matrix A. If A, B are congruent, i.e., B = P AP T for a nonsingular P , their cosquares are conjugate (e.g., see [9]). Lemma 7.2. Let β,γ F with βγ = 0. The subgroups Vβ = V (1, 1, 0,β) and ∈ 6 Vγ = V (1, 1, 0,γ) are conjugate in AGL3(F) if and only if β = γ. 1 Proof. Suppose that Q− Vβ Q = Vγ for some Q GL3(F) (see Proposition 3.4). We may assume that v Q = λv since v is the subspace∈ of F3 fixed pointwise by 1 1 h 1i both subgroups. Thus Q has shape: c 10 0 0 0 λ 0 0 p , p , Q = , λ =0, P = 1 1 1 2 nonsingular. 0 q1 p1,1 p1,2 6 p2,1 p2,2 0 q2 p2,1 p2,2 q1 q2 The matrix K = I E , E , normalizes Vβ. Hence, substituting Q with 4 − λ 3 2 − λ 4 2 1 1 1 1 KQ, we may suppose q = q = 0. Setting B = and C = we 1 2 0 β 0 γ have: 1 x1 X 1 y1 Y Vβ = 0 1 0 X = (x2, x3), Vγ = 0 1 0 Y = (y2,y3). T T 0 BX I2 0 CY I2 T T From VβQ = QVγ we get Y = XP and λBX = P CY , whence λBXT = P CP TXT, for all X F2. ∈ MORE ON REGULAR SUBGROUPS OF THE AFFINE GROUP 15 It follows that λB = P CP T, i.e., the matrices λB and C are congruent. So their cosquares must be conjugate. But the characteristic polynomials of the cosquares 2 1 2 1 are respectively t + (β− 2)t + 1 and t + (γ− 2)t + 1, whence β = γ. − − Lemma 7.3. Suppose d(Z(U)) = 2 and r(Z(U)) = 1. If U is abelian, then U = S(2,12). Otherwise, U is conjugate to exactly one of the following subgroups: 1 x1 x2 x3 01 0 0 (a) N1 = , if k(U)=2; 0 0 1 x1 00 0 1 1 x1 x2 x3 0 1 0 x2 (b) N2 = − , if k =1, d(U)=2 and char F =2; 0 0 1 x1 6 00 0 1 1 x1 x2 x3 0 1 0 x1 + x2 (c) N3,λ = , λ F∗, if k =1 and d(U)=3. 0 0 1 λx2 ∈ 000 1 The algebras 1= FI4 + N1, 2 = FI4 + N2 and 3,λ = FI4 + N3,λ have the following presentation:L L L 2 2 1 = SpanF(t1,t2), where t1 = t2 = t1t2 = 0; L 2 2 2 = SpanF(t1,t2), where t1 = t2 = t1t2 + t2t1 = 0; L 2 2 2 ,λ = SpanF(t ,t ), where t λt = t t = t t t =0, λ F∗. L3 1 2 2 − 1 2 1 1 − 1 2 ∈ Proof. If U is abelian, then r(U) = r(Z(U)) = 1 and by Lemma 5.3, U = S(2,12). So, suppose that U is not abelian. We may assume z = diag(J ,J ,J ) Z(U). 2 1 1 ∈ Now z = µ(v1) gives δ(v1) = 0 and, from (3), we get that the first row of δ(v) is 0 0 0 3 2 zero for all v F . Hence δ(v)= α y1 y2 with y + y2y3 = 0, by the unipotency β y y 1 ∈ 3 − 1 of U. Now fix v = (0, x2, x3) = 0 in v2, v3. Conjugating by g = diag(I2, P ) with a suitable P GL (F), we may6 assumeh eitheri (i) y = y = 0 and y =1 or(ii) ∈ 2 1 2 3 y1 = y2 = y3 = 0. 0 0 0 In case (i) if x3 = 0, we may suppose v = v2. So δ(v2)= α2 0 0 and δ(v3)= β2 1 0 0 0 0 α3 γ1 γ2 . Applying (2) to v2, v2 and to v3, v2 we obtain respectively α2 = 0 β γ γ 3 3 − 1 and γ1 = 1, γ2 = 0, which contradicts the unipotency of U. On the other hand, − −1 x3 0 if x = 0, conjugating by diag I , − − we may assume v = v , hence 3 2 x x 2 x 1 3 6 2 3 3 0 0 0 − 2 0 0 0 δ(v3) = α3 0 0 . Applying (2) to v3, v3 we obtain δ(v2) = δ(v3) = 0 0 0 . β3 1 0 α3 0 0 However, in this case U is abelian. In case (ii), up to a further conjugation by a matrix of the same shape of g, we 0 0 0 0 0 0 may suppose v = v2, hence δ(v2) = α2 0 0 . Set δ(v3) = α3 γ1 γ2 . Now, (2) β2 0 0 β3 γ3 γ1 − 2 applied to v2, v3 gives γ1δ(v2)+ γ2δ(v3) = δ(v2)δ(v3) = 0. In particular, γ2 = 0, 2 whence γ2 = 0. It follows γ1 = 0 by the condition γ1 + γ2γ3 = 0. Replacing, if necessary, v3 by a scalar multiple, we get either γ3 =1 or γ3 = 0. In the first case, (2) applied to v3, v3 gives α3 = β2, α2 = 0 and U is abelian. In the second one, U is conjugate V (α , α ,β ,β ) with α = β , from the non-abelianity. 2 3 2 3 3 6 2 16 M.A.PELLEGRINIANDM.C.TAMBURINIBELLANI Let ∆ = α2β3 α3β2, with α3 = β2. If ∆=0, then k(U) = 2. An isomorphism Ψ: = FI−+ V (α , α ,β ,β6 ) is obtained by setting L1 → L 4 2 3 2 3 Ψ(t1)= β2X2 α2X3 if α2 = 0 take − ; 6 Ψ(t2)= α3X2 α2X3 − β3 Ψ(t1)= α X2 + X3 if α2 = β2 = 0 take − 3 ; Ψ(t )= X 2 2 Ψ(t1)= X2 if α2 = α3 = 0 take β3 . Ψ(t2)= X2 + X3 − β2 By Proposition 3.4, the subgroup U is conjugate to N1. Suppose now that ∆ = 0 (which implies k(U)=1). If d(U) = 2, then char F = 2, α = β = 0 and β6 = α = 0. An isomorphism Ψ : = FI 6+ 2 3 2 − 3 6 L2 → L 4 V (α2, α3,β2,β3) is obtained by setting Ψ(t1)= X2, Ψ(t2)= X3. By Proposition 3.4, the subgroup U is conjugate to N2. If d(U) = 3, an isomorphism Ψ : 3,λ = FI4 + V (α2, α3,β2,β3) is obtained α2β3 α3β2 L → L by setting λ = (α −β )2 and 3− 2 Ψ(t1)= X3 if β3 = 0 take β3 α3 ; 6 Ψ(t2)= α β X2 + α β X3 − 3− 2 3− 2 Ψ(t1)= X2 if β3 =0, α2 = 0 take β2 α2 ; 6 Ψ(t2)= α β X2 + α β X3 − 3− 2 3− 2 α3 β2 Ψ(t1)= X2 + − X3 α3 if α2 = β3 = 0 (α3 = β2) take β2 . 6 − Ψ(t2)= X2 + X3 ( α3 β2 − − By Proposition 3.4, the subgroup V (α2, α3,β2,β3) is conjugate to N3,λ, λ = 0. The statement now follows from Lemma 7.2 6 We can now classify the regular subgroups U of AGL3(F) having linear δ, in- cluding the non-abelian ones, arising from d(Z(U)) = 2 and r(Z(U)) = 1. If d(Z(U)) = 4, then U is conjugate to S(4), by Lemma 5.2. If d(Z(U)) = 3, then U is one of the subgroups Rα described in Lemma 5.4: namely, for k(U) = 2, U is conjugate to R0 and, for k(U) = 1, U is conjugate to Rλ, where λ can be chosen in 2 a transversal F of (F∗) in F∗. As observed in Remark 6.4, R0 and R1 coincide, ♯ respectively, with S(3,1) and S(2,2). If d(Z(U)) = 2 we have two possibilities. When 3 r(Z(U)) = 2 we apply Lemma 7.1, that gives char F = 2 and U is conjugate to U1 . When r(Z(U)) = 1 we apply Lemma 7.3: when abelian U = S(2,12), otherwise U is conjugate to one of the subgroups N ,N ,N ,λ, λ F∗. 1 2 3 ∈ A complete set of representatives of the abelian regular subgroups of AGL3(F) is given in Table 2. The conjugacy classes of non-abelian regular subgroups of AGL3(F) are described in Lemma 7.3. 7.3. Case n =4. Once again, to obtain the full classification of the abelian regular subgroups of AGL4(F) we need some preliminary results. MORE ON REGULAR SUBGROUPS OF THE AFFINE GROUP 17 U FI4 + U char F Ψ Ker(Ψ) x0 x1 x2 x3 0 x0 x1 x2 4 S(4) any (19) t1 indec. 0 0 x0 x1 h i 0 0 0 x0 x0 x1 x2 x3 0 x0 x1 0 3 2 S(3,1) any (13) t1 ,t2 ,t1t2 0 0 x0 0 h i 0 0 0 x0 x0 x1 x2 x3 Rλ, λ F 0 x0 0 x1 2 2 ♯ any (15) t1 λt2 ,t1t2 indec. R S∈ ( 1= (2,2)) 0 0 x0 λx2 h − i 0 0 0 x0 x0 x1 x2 x3 3 0 x0 0 x2 2 2 U1 2 (22) t1 ,t2 indec. 0 0 x0 x1 h i 0 0 0 x0 x0 x1 x2 x3 0 x0 0 0 2 S(2,12) any (19) t1,t2,t3 0 0 x0 0 h i 0 0 0 x0 Table 2. Representatives for the conjugacy classes of abelian reg- ular subgroups U of AGL3(F), for any field F. Lemma 7.4. Let U be a regular subgroup of AGL4(F) such that δ is linear. If d(Z(U)) = r(Z(U)) =3, then U is abelian and is conjugate to 1 x1 x2 x3 x4 0 1 x1 0 x3 R(α, β)= 0010 0 , α,β F. ∈ 0 0 βx3 1 x1 + αx3 0000 1 Furthermore, if F has no quadratic extensions, there are exactly two conjugacy classes of such subgroups, whose representatives are, for instance, R(0, 0) and 4 R(1, 0), which is conjugate to S(3,2). Finally, U1 = R(0, 0) is indecomposable. Proof. We may suppose that z = diag(J ,J ) Z(U). From z,z2 U we obtain 3 2 ∈ ∈ δ(v1)= E1,2 + E3,4 and δ(v2) = 0. By Lemma 4.1 an the unipotency of U it follows that δ(v3)= E1,4 +αE3,4 +βE3,2 +γ(E3,1 +E4,2). Applying (2) to v3, v1 we obtain 3 δ(v4)= δ(v3)δ(v1)= γE3,2. In particular U is abelian. Now (µ(v3) I5) = 0 gives γ = 0. We conclude that U is conjugate to R(α, β). − Now, assume that F has no quadratic extensions. An epimorphism Ψ : F[t1,t2] 2 → FIn + R(α, β) is obtained in the following way. If α +4β = 0, take +1 6 (24) Ψ(t ) = (α + α2 +4β)X 2X , Ψ(t ) = (α α2 +4β)X 2X 1 1 − 3 2 − 1 − 3 when char F = 2 andp take p 6 (25) Ψ(t1)= rX1 + X3, Ψ(t2) = (r + α)X1 + X3, 18 M.A.PELLEGRINIANDM.C.TAMBURINIBELLANI when char F = 2 (here r F is such that r2 + αr + β = 0). In both cases, Ker(Ψ) = t 3,t 3,t t . Comparison∈ with the presentation of FI + S given h 1 2 1 2i 5 (3,2) in (18) shows that R(α, β) is conjugate to S(3,2) by Proposition 3.4. If α2 +4β = 0, take (26) Ψ(t )= X , Ψ(t )= αX 2X 1 1 2 1 − 3 when char F = 2 and take 6 (27) Ψ(t1)= X1, Ψ(t2)= βX1 + X3 3 2 2 when char F = 2. In both cases Ker(Ψ) = t1 ,tp1 t2,t2 . The algebras defined by the two presentationsh abovei are not isomorphic, as the 2 2 subspaces consisting of elements whose square is zero (namely t1 ,t2 in the first 2 h i case, t1 ,t2,t1t2 in the second case) have different dimensions. By Proposition 3.4, thereh are exactlyi two conjugacy classes of subgroups R(α, β), depending on the nullity of α2 +4β. 4 4 Noting that k(U1 ) = 2, direct computation shows that U1 is indecomposable. Lemma 7.5. Let U be an abelian regular subgroup of AGL4(F). If d(U)=3 and r(U)=2, then U is conjugate to 1 x1 x2 x3 x4 0 1 x 0 0 1 R(α,β,γ)= 00 1 00 . 0 0 αx3 + βx4 1 0 0 0 βx3 + γx4 0 1 If β2 αγ = 0, then k(U)=1. If β2 αγ = 0 with (α,β,γ ) = (0, 0, 0), then k(U)=2− . Finally,6 if α = β = γ =0, then− k(U)=3. Furthermore,6 assuming that every element of F is a square, there are exactly three conjugacy classes of such subgroups. Their representatives are, for instance, 4 U2 = R(0, 0, 1) (if k = 1), R(0, 1, 0) (if k = 2), R(0, 0, 0) = S(3,1,1) (if k = 3). ♯ Observe that R(0, 1, 0) is conjugate to S(2,2,1). Proof. We may assume z = diag(J ,J ,J ) U. From z,z2 U we obtain δ(v )= 3 1 1 ∈ ∈ 1 E1,2 and δ(v2) = 0. By Lemma 4.1 and the unipotency of U, for any v v3, v4 000 0 ∈ h i 000 0 2 we have δ(v)= 0 α ξ η , where ξ + ηǫ = 0. Now, rk(µ(v1 + v3 + v4) I5)=3 0 β ǫ ξ − − implies δ(v3) = α3E3,2 + β3E4,2 and δ(v4)= α4E3,2 + β4E4,2. Since U is abelian, (3) applied to v3, v4 implies α4 = β3. Hence U is conjugate to R(α,β,γ). Suppose first that ∆ = β2 αγ = 0. In this case k(U)=1. If char F = 2 define − 6 6 if α = 0 if α =0 6 Ψ(t1) = X1, Ψ(t1) = βX1, (28) β+√∆ α γ Ψ(t2) = X3 X4, Ψ(t ) = X + βX , 2∆ − 2∆ 2 − 2 3 4 β+√∆ Ψ(t3) = X3, Ψ(t3) = − α X3 + X4, and if char F = 2, define Ψ(t ) = √4 ∆X + √γ X + √α X , 1 1 √4 ∆ 3 √4 ∆ 4 (29) Ψ(t2) = √αX1 + X3, Ψ(t3) = √γX1 + X4. MORE ON REGULAR SUBGROUPS OF THE AFFINE GROUP 19 2 2 2 We get Ker(Ψ) = t1 t2t3,t2 ,t3 ,t1t2,t1t3 . Next suppose β2h αγ− = 0 and (α,β,γ) = (0i , 0, 0). Then k(U) = 2. Define − 6 if α = 0 if α = 0 and γ =0 6 6 Ψ(t ) = X , (30) 1 1 Ψ(t1) = √γX1, Ψ(t ) = β√α+1 X + αX , Ψ(t ) = X , 2 − √α 3 4 2 4 Ψ(t ) = βX + αX , Ψ(t3) = X3. 3 − 3 4 2 2 2 In both cases, Ker(Ψ) = t1 t2 ,t3 ,t1t2,t1t3,t2t3 . Comparison with the pre- ♯ h − i ♯ sentation of FI5 + S(2,2,1) given in (20) shows that U is conjugate to S(2,2,1). Finally, if α = β = γ = 0, then R(0, 0, 0) = S(3,1,1) and k(U) = 3. Remark 7.6. Consider the algebra of Example 3.2 and the corresponding regular subgroup R = R( 1, 0, 1) of theL previous Lemma. Since k(R) = 1, under the assumption that −1 is a− square in F, we obtain that the algebras − F[t ,t ,t ] F[t ,t ,t ] 1 2 3 and 1 2 3 t 2 + t 2,t 2 + t 2,t t ,t t ,t t t 2 t t ,t 2,t 2,t t ,t t h 1 2 1 3 1 2 1 3 2 3i h 1 − 2 3 2 3 1 2 1 3i are isomorphic in any characteristic. Actually, if char F = 2, an isomorphism can also be obtained directly via the change of variables t1′ = t1 + t2 + t3, t2′ = t2 + t3, t3′ = t1 + t3. This fixes an inaccuracy of [12], corrected in [13]. Lemma 7.7. Let U be an abelian regular subgroup of AGL4(F). If d(U)= r(U)= 2, then char F =2 and U is conjugate to U 3 S , where U 3 is defined in Table 2. 1 × (1) 1 Proof. We may suppose z = diag(J2,J2,J1) U. From z U we get δ(v1)= E2,3. 2 ∈ ∈ Lemma 4.1 and the condition (µ(v2) I5) = 0 give δ(v2) = E1,3 + α2E4,3. Now, 2 − (µ(v1 + v2) I5) = 0 implies char F = 2. Applying (2) to v2, v1 we get δ(v3) = − 2 δ(v2)δ(v1) = 0. From(µ(v2+v4) I5) = 0 we obtain δ(v4) = (α2+α4)E2,3+α4E4,3, but (µ(v ) I )2 = 0 gives α =− 0. In conclusion, U is conjugate to 4 − 5 4 1 x1 x2 x3 x4 0 1 0 x 0 2 R(α)= 0 0 1 x1 + αx4 0 . 000 1 0 0 0 0 αx2 1 An epimorphism Ψ : F[t ,t ,t ] FIn + R(α) is obtained by setting 1 2 3 → +1 (31) Ψ(t1)= X1, Ψ(t2)= X2, Ψ(t3)= αX1 + X4. 2 2 2 3 We have Ker (Ψ) = t1 ,t2 ,t3 ,t1t3,t2t3 . Considering the presentation of U1 given in Table 2, we haveh that U is conjugatei to U 3 S . 1 × (1) We now classify the abelian regular subgroups U of AGL4(F). If d(U) = 5, then U is conjugate to S(5) by Lemma 5.2. If d(U) = 4, then by, Lemma 5.4, U ♯ is conjugate to R0 = S(4,1) when k(U) = 2, and to R1 = S(3,2) when k(U) = 1. 4 Suppose d(U)=3. If r(U) = 3, then U is conjugate either to U1 or to S(3,2) by 4 ♯ Lemma 7.4. If r(U) = 2, then U is conjugate either to U2 or to S(2,2,1) or to S(3,1,1) by Lemma 7.5. Finally suppose d(U) =2. If r(U) = 1 then U is conjugate to 3 S(2,13) by Lemma 5.3. If r(U) = 2, then char F = 2 and U is conjugate to U1 S(1) by Lemma 7.7. × When F has no quadratic extensions, a complete set of representatives of the conjugacy classes of abelian regular subgroups is given in Tables 3 and 4. 20 M.A.PELLEGRINIANDM.C.TAMBURINIBELLANI U FI5 + U char F Ψ Ker (Ψ). x0 x1 x2 x3 x4 0 x0 x1 x2 x3 5 S 0 0 x0 x1 x2 any (19) t (5) h 1 i 0 0 0 x0 x1 0 0 0 0 x0 x0 x1 x2 x3 x4 0 x0 x1 0 x2 ♯ 3 2 S 0 0 x0 0 x1 any (21) t t ,t t (3,2) h 1 − 2 1 2i 0 0 0 x x 0 3 0 0 0 0 x0 x0 x1 x2 x3 x4 0 x0 x1 0 x3 =2 (24) U 4 0 0 x 0 0 6 t 3,t 2t ,t 2 1 0 h 1 1 2 2 i 0 0 0 x0 x1 2 (25) 0 0 0 0 x0 x0 x1 x2 x3 x4 0 x0 0 0 x1 =2 (28) 4 6 2 2 2 U 0 0 x0 0 x3 t t t ,t ,t ,t t ,t t 2 h 1 − 2 3 2 3 1 2 1 3i 0 0 0 x0 x2 2 (29) 0 0 0 0 x0 Table 3. Representatives for the conjugacy classes of indecom- posable abelian regular subgroups U of AGL4(F), when F has no quadratic extensions. Acknowledgments. 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MORE ON REGULAR SUBGROUPS OF THE AFFINE GROUP 21 U FI5 + U char F Ψ Ker(Ψ) x0 x1 x2 x3 x4 0 x0 x1 x2 0 4 2 S 0 0 x0 x1 0 any (19) t ,t ,t t (4,1) h 1 2 1 2i 0 0 0 x0 0 0 0 0 0 x0 x0 x1 x2 x3 x4 0 x0 x1 0 0 3 3 S 0 0 x0 0 0 any (21) t ,t ,t t (3,2) h 1 2 1 2i 0 0 0 x x 0 3 0 0 0 0 x0 x0 x1 x2 x3 x4 0 x0 x1 0 0 S 0 0 x 0 0 any (19) t 3,t 2,t 2,t t ,t t ,t t (3,1,1) 0 h 1 2 3 1 2 1 3 2 3i 0 0 0 x0 0 0 0 0 0 x0 x0 x1 x2 x3 x4 0 x0 0 x1 0 ♯ 2 2 2 S 0 0 x0 x2 0 any (21) t t ,t ,t t ,t t ,t t (2,2,1) h 1 − 2 3 1 2 1 3 2 3i 0 0 0 x0 0 0 0 0 0 x0 x0 x1 x2 x3 x4 0 x0 0 x2 0 3 2 2 2 U S 0 0 x0 x1 0 2 (31) t ,t ,t ,t t ,t t 1 × (1) h 1 2 3 1 3 2 3i 0 0 0 x0 0 0 0 0 0 x0 x0 x1 x2 x3 x4 0 x0 0 0 0 2 S 3 0 0 x0 0 0 any (19) t ,t ,t ,t (2,1 ) h 1 2 3 4i 0 0 0 x0 0 0 0 0 0 x 0 Table 4. Representatives for the conjugacy classes of decompos- able abelian regular subgroups U of AGL4(F), when F has no qua- dratic extension. [12] B. Poonen, Isomorphism types of commutative algebras of finite rank over an algebraically closed field, in Computational arithmetic geometry, 111–120, Contemp. Math., 463, Amer. Math. Soc., Providence, RI, 2008. [13] http://www-math.mit.edu/∼poonen/papers/dimension6.pdf, 11 January 2016. [14] M.C. Tamburini Bellani, Some remarks on regular subgroups of the affine group, Int. J. Group Theory 1 (2012), no. 1, 17–23. Dipartimento di Matematica e Fisica, Universita` Cattolica del Sacro Cuore, Via Musei 41, I-25121 Brescia, Italy E-mail address: [email protected] Dipartimento di Matematica e Fisica, Universita` Cattolica del Sacro Cuore, Via Musei 41, I-25121 Brescia, Italy E-mail address: [email protected]